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Wednesday, May 31, 2006

Geometric Sequences / First Day

Hey guys, it's Manny doing the scribe for today, cause yeah Michael couldn't so I volounteered for it. Today, we started our new unit Geometric Sequences, and thankfully, or sadly for some, it's our last. It was a lot to absorb without taking some form of notes, so I'll try my best to describe it to you guys.

So what is a sequence? A sequence is a list of numbers that follows a certain rule. That's what it is.

In the beginning of class, Mr.K put up a question with 6 parts that looked like this:

Find the next Four Terms in Each Sequence:

a) 4, 7, 10, 13, __, __, __, __

b) 3, 6, 12, 24, __, __, __, __

c) 1, 2, 4,__, __, __, __

d) 2, 5, 10, 17, 26, __, __, __, __

e) 2, 1, 3, 4, 7, 11 __, __, __, __

f) 77, 49, 36, 18, __, __, __, __

(We talked a lot about the first two.)

Well, at first we thought it was real easy, something done in like grade five. The first one (part a) we said, add 3 to the last number, easy as cutting butter. Well it was, but why is it 3? We all went speechless. Then we said that because taking two that are beside eachother and finding the difference between them will give us 3, so therefore it always has to go up by 3. So this is what we did.

7 - 4 = 3
10 - 7 = 3
13 - 10 = 3
And it just so happens that the 3 is called the common difference.

So this is what we have so far for part a) 4, 7, 10, 13, 16, 19, 22, 25.

Then, Mr. K, asked so what would the 11th number be? We answered 34. It was right, and we did this by adding 3, to the last number three more times, since there are eight terms already.

It turns out Mr.K made a "mistake," he meant to ask for the 1001th number. Well the 11th, and 1001th, or n in this case, are called ranks. One of us saw that it was one number less than the rank, multiplied by 3 and added to the first terms which was 4. The term is the output of the rank.

We came up with the formula tn = t1 + d(n-1).
Where tn is the term of the number (or rank number),and Where t1 is the first term, and
Where d is the common difference.
So therefore, tn = 4 + 3(n-1).
And for the problem for the 1001th looks like this t1001 = 4 + 3(1001-1), which equals out to a huge number bigger than googal =] (There is not as much atoms in the universe than the number googal 10100).
This type of equation has a name called the Recursive Definition, because recursive means to apply some feature over and over again, here we are adding 3 all the time.

Well now that we know all this information, what is the initial term, Mr. K asks? Well since we added 3 to the last number to get the next, we can just subtract 3 from the first. And at the 0th rank, we have 1. This is what it looks like.

rank (n)012345678...
term (t)147101316192225...

Since we now know the 0th term, we can now write another equation. And come up with tn = dn + to, therefore being tn = 3n + 1. This equation is called the Impicit Definition, because implicit means that is implied, not obvious, and is in the subtext kind of like when some of us nodd or a raise an eyebrow to say hello to a friend while an adult would not understand it.

Then we graphed it to look like this.


We did this to show that from this pattern, given only 4 pieces of info, we can show it
symbollically - written as an equation
numerically - table of values
graphically -drawn out as a graph
Which is shown above. These are the 3 perspectives you can look at just about anything given at least one of them.

We are only learning about two types of sequences in this course. Arithmetic, and geometric. Arithmetic deals in adding/subtracting types of patterns, which was what we just did. Geometric is dealing with multiplication/division types of patterns.

Part b) 3, 6, 12, 24, __, __, __, __, is geometric.

- Will explain soon -

If anyone is curious the answers to the question above are:

a) 4, 7, 10, 13, 16, 19, 22, 25 [common difference 3]

b) 3, 6, 12, 24, 48, 96, 192, 384 [common ratio 2]

c) 1, 2, 4, 8, 16, 32, 64 [powers of 2 raised to the exponent of the (rank - 1)]

d) 2, 5, 10, 17, 26, 37, 50, 55, 72 [square the rank + 1]

e) 2, 1, 3, 4, 7, 11, 18, 29, 37 [sum of two side by side numbers, or fibonacci sequence]

f) 77, 49, 36, 18, 8, 8, 8, 8 [multiply digits]
or
77, 49, 36, 18, 08, 0, 0, 0 [multiply digits, but have to have 2 digits]

We only did the first two today, and will not be looking at the last three parts, as they are different from arithmetic and geometric sequences.

Well, I have had a long day today. I was asked to volounteer last minute, and from the kindness of my heart I couldn't refuse, just like how I volounteered for this scribe today. But I am tired, and starving, I have had nothing but junk food since the past 14 hours, and it's way past my bedtime. I will continue the post tomorrow morning, and if not by the end of tomorrow. I'm sorry if this causes an inconvience to anybody and will be happy to fill you in if you want face to face.

The scribe for tomorrow's class will I guess be Michael since he didn't do it today, and if that's not possible, it'll be Master Teddie ;).



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Tuesday, May 30, 2006

blogging for the test!

Overall this unit wasn't all that bad. I guess the thing that confuses me the most is questions dealing with the base law, and comparing it to another base. Also what I have a little trouble on is gathering all the information from a long answer question, to help solve the problem. Well I guess that's about it, goodluck everyone!



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blogpost

the unit exponents and logarithms is less confusing to me than the last unit which is probability. i like this unit except on one part. the logarithm expression raised on an exponential expression often made me confused especially when a logarithm expression is raised on another logarithm expression. i get it after mr. k explains it to us but i forgot how it happen again after a while. well, that's my only big problem about this unit.



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blogging on blogging

Well this unit is going well for me. No bumpy roads except for the word problems, namely the ones with the least amount of information. Everything else I could do nicely. I don't have anything else to say except for the need of more word problems for practice



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Rap-up of Exponents and Logarithm

My last scribe ever...YAY! I'm pretty happy about that but I'm also disappointed that my own right hand man *ahem Jefferson betrayed me! So much for taking over the blog. Anyways I guess I'm on my own now and I should get right to it since we did so much work today, so here it goes...

In the morning class we got right down to business. Mr. K put up old exam questions on the board. There were two parts, a calculator and a non calculator part. The first part you were allowed to use a calcualtor:

1) 3x+4 = 72x+1
2) If loga2=0.3562 and loga5=0.8271 show that loga40=1.8957
3) A new car costs $24 000. It's value after t years is given by V=24000(0.8)t
a.) Determine the value after 8 years.
b.) How many years will it take for its value to decrease to one-eighth of its initial value?

For the rest of the questions calculators weren't allowed:

4) A population triples every 7 days. Determine an expression (formula) for the total population after t days if the initial population is 800.
5) Evalute: log(100 sin(π/2))
6) State the range of f(x)=2-x
7) Given:

1/2 ≤ log x ≤ 2

find the minimal and maximum value of x.

Solutions:

1)      log3x+4 = log72x+1          ← change it as a log
(x+4)log3 = (2x+1)log7 ← by the rules of logarithm
xlog3+4log3 = 2xlog7+log7 ← distribute the brackets because the brackets are #'s
xlog3-2xlog7 = log7-4log3 ← get all the #'s and x's together
x(log3-2xlog7) = log7-4log3 ← factor out the x
x = (log7-4log3) ← solve for x using a calculator
(log3-2log7)
x = 0.8766

2) loga40 = loga(23∙5)
= loga23+loga5
= 3loga2+loga5
= 3(0.3562)+(0.8271)
= 1.0686+0.8271
= 1.8957


3.a) V(8)=24000(0.8)8
= 4026.5318
3.b) (1/8)24000 = 3000
original value=24000 final=3000 therefore,

3000 = 24000(0.8)t
24000 24000
(1/8) = (0.8)t
ln(1/8)=tln(0.8)
ln(1/8)=t
ln(0.8)
9.3189 ≈ t ← using a calculator evaluate t


4) P =7days           A=A°Mt/p
M =3 A=800(3)t/7 ← just substitute the values
A° =800
t =?

5) log(100 sin(π/2))
=log(100∙1) ← since sin(π/2)=1
=log(100)
=2

6) f(x)=2-x
So to solve this question we must think back on transformations.
We know that this is an exponential decay because the basic function (2x) is an
exponential growth. Now in transformations when x is negative that means it "flips
over the y-axis" in other words the x-coordinates switch.
Therefore the answer must be (0,∞) because it only approaches 0 but never touches it.

7) 101/2 ≤ 10logx ≤ 102 ← changing it to base 10 will simplify it
√10 ≤ x ≤ 100


Everyone was kind of stuck on that last question so Mr. K
tried to clear it up by giving us another question which
he thought was mildy amusing.
log(logx)=2
102=logx
100=logx
10100=x which we know as Googol!
humorous? i didnt think so.


Now that took the whole morning class and in the afternoon class we wrote in our dictionary for the whole period. There were five pages of notes so for those students that missed class or didn't finish copying here it is:

PROOFS OF THE LAWS OF LOGARITHMS

Proof of the the product laws
Prove logaMN = logaM + logaN
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
x=logaM IFF ax=M
y=logaN IFF ay=N

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaMN=loga(ax∙ay)
=logaa(x+y)
=x+y
=logaM+logaN
Q.E.D


Proof of the quotient law
Prove loga(M/N)=logaM-logaN
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
x=logaM IFF ax=M
y=logaN IFF ay=N

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
loga(M/N)=loga(ax/ay)
=logaa(x-y)
=x-y
=logaM-logaN
Q.E.D


Proof of the power law
Prove logaMc=clogaM
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaM=b IFF ab=M

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaMc=loga(ab)c
=logaacb
=cb
=clogaM
Q.E.D


Proof of the base law
Prove:
logaMc= logbM
logba
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaM=c IFF ac=M

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
ac=m
logbac=logbM
clogba= logbM
logba
c=logbM
logba
logaM= logbM
logba
Q.E.D


EXPONENTIAL MODELING

the basic function:

How we model a real life situation depends on how much information we are given.

case 1: Given a minimal amount of information (A, A°, ▲t) we will create a model in base 10 or base e. (base e is prefered)
A is the amount at the end of the time period.
A° is the original amount of the "substance".
MODEL is an exponential function that describes the growth (or decay) of the "substance".
t is the time required for the amount of the "substance" to change from A° to A.

NOTE: Some "substances" are populations. The model is usually an exponential function of the form 10k, or er where k and r are constants.

EXAMPLE: The population of earth was 5 billion in 1990. In 2003 the population grew to 6.3
a) Model the population as an exponential function.
b) What was the population of earth in 1967?


a)                let t=o in 1990.
A°=5 A=A°(model)t
A=603 (in 2003) 6.3=5(model)13
t=2003-1990 6.3 =(model)13
=13 5
From here you can go on in two different ways. It wouldn't really matter which way you choose because it will end up having the same answer.

if you choose base 10                            if you choose base e
log(6.3/5)=log(model)13 ln(6.3/5)=ln(model)13
log(6.3/5)=13log(model) ln(6.3/5)=13ln(model)
1/13log(6.3/5)=log(model) 1/13ln(6.3/5)=ln(model)
0.0077=log(model) 0.0178=ln(model)
100.0077= model e0.0178= model
therefore P=5(100.0077t) P=5e0.0178t

b) If t=0 in 1990 then t=-23 in 1967.
if base 10 if base e
P=5(100.0077(-23) P=5e0.0178(-23)
P=3.3219 P=3.3219


case 2: Given lots of information (A, A°, M, p)A is the amount at the end of the time period t.
A° is the original amount.
M is the "multiplication factor" or growth rate.
t is the time has passed.
p is the time period; time required to multiply by M once.

EXAMPLE 1: A colony of bacteria double every 6 days. If there were 3000 bacteria at the start of an experiment how many bacteria will there be in 15 days?
 A=?             A=A°Mt/p
A°=3000 =3000(2)15/6
M=2 =16970.5628
t=15 ≈16971 bacteria
p=6


EXAMPLE 2: The half life of carbon 14 is 5700 years. How much of a 10mg sample will remain after 4500 years?
 A=?             A=A°Mt/p
A°=10 =10(1/2)4500/5700
M=1/2 =5.8022 mg
t=4500
p=5700
And that's that!! Phew...three five small coffee cups later and I'M FINALLY DONE! YAY! Okay so now that I'm all finished and I'm in a cranky mood I will pick none other than michael as the scribe for tomorrow. Nah, enjoy!



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Logs and Exponents Online Quiz

The quiz goes online at 3:30 this afternoon. The deadline is 9:00 am Thursday morning. The unit test will be on Friday.

You can write the quiz here. You have 45 minutes to complete it. If you haven't yet registered at QuizStar then when you get to the site follow these instructions:

  1. Click on the big yellow "Sign Up" arrow.

  2. Use only your first name and last initial as indicated.

  3. Pick a username that will allow me to easily identify you, i.e. first name and last initial.

  4. Make up any password you like.

  5. Click on [Register] then [Search] by teacher's name (kuropatwa) and you'll find me.

  6. Click on the box next to Pre-Cal 40S and then [Register].

  7. Follow the instructions on the screen.


Actually, if you read each page carefully, you'll see that the sign up process is very straight forward and self explanatory. If you hit any snags email me and we'll sort it out together.

Remember, the quiz is timed. You'll only have 45 minutes to complete it once you've begun. It consists of 11 multiple choice questions (1 mark each). DON'T PANIC. Take your time. I know you'll all do well. ;-)

Do Your Best!



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Sisyphus' Bloggon the Blog 7

Logarithms and exponents. A logarithm is an exponent. This unit isn't too bad, well not as bad as the time during the unit counting. I understand most of the simple short answer questions. The thing that I have trouble with, as do probably most other classmates, are the long answers. Yesterday's class doing the group work, was tough! We did not get a single question right. We just didn't understand the question. Fortunately a few groups were lucky to fluke the answer, but does not make sense to the problem. I think you should drop that assignment Mr.K ;). So, like the other units... It's those darn long answers!



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Monday, May 29, 2006

Jefferson's FINAL SCRIBE - ACTION ON EXPONENTS AND LOGARITHMS!!

HEY HO! Its scribe specialist and former scribe emperor, Jefferson presenting today's scribe post! Attention! The logarithm test will be on Friday, the wiki project will also be due on Friday.

Mr.K started off today's lesson with a quiz. For those who were absent here is a copy of the quiz:



Now lets go through the quiz together shall we?

(1) Find log10(0.001)

solution:

log10(0.001) = -3

(2) Find log2(64) + log3(9)

solution:

log2(64) + log3(9) = 6 + 2
log2(64) + log3(9) = 8

(3) 3x = 7; which is true?

(a) 7= logx(3) (b) x = log7(3) (c) 3 = logx(7) (d) 7 = log3(x) (e) x=log3(7)

solution:
REMEMBER a log is an exponent. "7" is the power so the log format would be:
logbasepower = exponent
so the answer would be (e) x=log3(7)

(4) if loga(25) = 4, what is a?

(a) 1/5 (b) sqr(5) (c) 5 (d) log2 (5) (e) none of these

Loga(25) = 4
a = sqr(5)

(5) find log100(1 000 000)

log100(1000000) = 3

(6) If log10 2 ≈ 0.301, which of these is the closest to the log10(2000)?

(a) 0.6 (b) 3 (c ) 6 (d) 30 (e) 60 (f)6

log10(2000) = log10(2 x 1000)
log10(2000) = log10(2) + log10 (1000)
log10(2000) = 0.301 + 3
log10(2000) = 3.301


so the answer is (b) 3

(7) If log10 2 ≈ 0.301, which of these is the closest to log10(8)?

(a) 0.3 (b) 0.6 (c )0.9 (d) 1.2 (e) 2.4 (f) 6

log108 = log10(2)3
log108 = log10(2)3
log108 = 3log10(2)
log108 = 3(0.301)
log108 = 0.903


and the answer is (c) 0.9

(8) If log10 2 ≈ 0.301, which of these is the closest to log2(10)?

(a) 0.5 (b) 1 (c ) 3 (d)5 (e)20 (f)50

log2(10) = log2(2*5)
log2(10) = log2(2) + log2(5)
log2(10) = 1 + 2.32
log2(10) = 3.32


and the answer is (c) 3

(9) find 2log2(17)

solution:

17

(10) find 4log2(3)

4log2(3) = 9
(22)log2(3) = 9
22log2(3) = 9
Log2(3)2 = 9
2log2(9) = 9


*not sure if this is correct Mr k , or anyone else please correct help me correct this*

(11) if log3((10) = k, then log9(10) =

(a) 2K (b) k/3 (c )k/2 (d)k2 (e)2.4 (f)sqr(K)


log3(10) = K
3K =10


*since we don't know what the log9(10) is we make it equal to any variable we name, lets call it "n" courtesy of Mr k.*

log9(10) = n
9n = 10


3K = 9n
3K = 32n
K = 2n
k/2 = n

and the answer is k/2

After the quiz, we were put into groups and faced a difficult word problem. Again, for those who were absent, here is a copy of the hand out:



AND NOW LETS GO OVER THE QUESTIONS!!

(a) The intensity of thunder is 1012 times the intensity of barely audible sound. What is the decibal level of thunder?

I = 1012 * 10-12
I = 1

* I = intensity of thunder * barely audible sound *

D = 10log(1/10-12)
D = 10log1012
D = 120

(b) If the decibal levels of a subway train entering a station and normal conversation are 100 dB and 60 dB, respectively, how many times as intense as normal conversation is the noise of the subway train?

Dt = 100
100 = 10log(I * 1012)
10 = logI + log1012
-12 + 10 = logI + logI
-2 = logI
10-2 = It

Dc = 60
60 = 10log(1012 * I)
6 = log(10)12 + logI
6 = 12 + logI
-12 + 6 = logI
-6 = logI
10-6 = log In

It/In = 10-12/10-6
= 104
=10000 times more intense

*I had just resently obtain the last bit of the answers after class, unfortunately I do not understand what's going on. Hopefully Mr K will explain this solution when he sees this scribe*

WELL I AM DONE!! Took me way too long to do this scribe. Operation rebellion will commence by a new member of the anti scribe take over movement, namely me =). My first victim will be NONE OTHER THAN JANET!!
that is all, *runs away*

later days, Jefferson








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Sunday, May 28, 2006

3D Tic-Tac-Sunday



Better late than never. ;-)

I missed the Sunday Funday post last week and I really don't want to make that a habit so this week I'm posting two versions of this week's game ... 3D Tic-Tac-Toe. Here are the instructions for the game pictured on the left (may take a little while to load) ...

The object of 3D Tic Tac Toe is to get four in a row horizontally, vertically or diagonally on one plane or across all four planes.

You are Red, the Computer is Blue. In the first game, you go first. In subsequent games, the loser goes first. If the game ends in a tie, whoever went first will do so again.

The moves of the game are notated on the right. Moves that threaten a win are noted with an *. The Computer plays a strong, but not unbeatable game. Good luck!!!


These are the instructions for the game on the right ...

This game is basically a 3D tic-tac-toe, except that the rules have been changed to keep the first player from winning all the time. Here you have to form two rows that meet at a right angle (in the shape of an "L").


Have Fun!!



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Blogging on blogging

I like this unit because it is more algebra than the previous unit that we have. Maybe, analyzing the word problem is giving me a hard time so far. However, more practice I think I can handle it on our test.



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BLOG ON LOG

The logarithms and exponents unit was a real challenge. Mainly because of the new concepts and new ways to look at fuctions in terms of exponents. It takes some time to get your head around it but eventually i get it. And I'm kinda confused in what ln is and wats the purpose of it.



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Saturday, May 27, 2006

Last Scribe of year

Today, acually friday, we had a sub for both classes.

In the morning we worked in groups on a group assignment, about Logarithm word problems.
The question on the assignment was:

Sugar is put into a large quantitiy of water and the mixture is stirred. After 2 minutes 50% of the sugar had dissolved. How much longer will it take until 90% of the sugar has dissolved (Model the sugar dissolving as an exponential decay function.) ?

What we know :
Original = 100%
This is 100% because when we begin the experiment none of the sugar has dissolved, therefore there was still 100% in the mixture.
Final = 10%
This is 10% because we want to know how much time it will take 90% of sugar to become dissolved leaving 10% of the sugar left.
Life = 1/2
This is because we know that after every 2 minutes of the experiment 50% of the sugar will be dissolved which is also known as a half-life.

With that information we plug it into the equation:
0.1 = 1 ( 1/2) t/2

0.1 Represents the the final which is 10%
1 Represents the initial which is 100%
1/2 Represents the life which is 50%
2 Represents the amount of time it will take for the life to decay

0.1 = 1 ( 1/2) t/2
0.1 = (1/2) t/2
ln 0.1 = (t/2) (ln 1/2)
(ln 0.1) / (ln 1/2) = t/2
2 ((ln0.1) / (ln 1/2)) = t
6.6439 = t
It will take approximately 7 minutes for 90% of the sugar to become dissolved in water.

*NOTE: I am not sure if this is the right answer as we did not go over it in class, but it is what the majority of the class got so i'm assuimg it is.

After the group assignment we did some questions from out text book for the rest of the period.
Pg. 83 Questions 30 - 34

In the afternoon we just did questions in our text book for the period.
Pg. 119 Questions 1 - 33 Odd

That was all we did for friday's class. Last scribe of the year and I get a not so bad one =D. Anyway monday's scribe is Jefferson haha who better carry out our plan for next week =P.



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Friday, May 26, 2006

B.O.B.

LOGARITHMS . . . .umm I sort of get it. Long answer questions are somewhat complicated though because I don't know which equation to use and what to do. This unit was really short and so I found it hard to absorb all the log information. Hopefully we'll practice some long answer questions? Well there isn't much to say about this unit, hopefully the test won't be hard? haha I doubt it. Later math people.

=) =) =) =) =) =) =)



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Thursday, May 25, 2006

LAST SCRIBE OF THE CYCLE

Today we had one class for today. We started off by writting down notes in our Math Dictionaries.

The Compound Interest Formula


A is the amount at theend of the interest period.

P (A.K.A. Ao) is the principle amount originally invested.

r is the percent rate of return. Written as a decimal.

n is the number of times the principle is compounded each year.

t is the time in years.

Interest Compounded Countinuously


A is the amount at the end of the investment period.

P is the principle amount originally invested.

r is the percent rate of return. Written as a decimal.

t is the time in years.

That was all the notes we had written down. Next, he wrote down some examples then problems.

Examples

C14 has a half life of 5700 years.

Some bones that have been found only have 10% of C14 that they should have. How old are the bones?


A colony of wasps has 800 members. Every 8 days the colony triples in size. How many wasps will there be in the colony in 3 weeks?



There will be 14306.

Solve

Radon has a half life of 3.8 days. How much of a 5g sample will remain after 4.2 days?



The population of a colony of pre-cal 40s students grew from 3x105 to 4x105 during the period between noon and 2pm.At what time will the population be at 6x105?



Well that's all that we did for class today. Now the next scribe will be . . . . . . . ABRIEL muahahahhahahahahah ;D

DON'T FORGET TO DO YOUR HOMEWORK. . PAGE 82 #24-29

thats all folks




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blogg on the logs

this topic seemed pretty easy, the only thing i need to learn and pac down currently is a number to the power of a log, that stuff kind of doesnt "click" in my brain. other than that, i think i should do pretty good on the test, im aiming for success!



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Wednesday, May 24, 2006

May 24th Scribe By Marcquin

Well lets see here... Todays first class in morning we began with some questions.

Write as an expontial
a) log264 = 6
64 = 26

b) log7 343 = 3
343 = 73

Write as a logarithm
a) 38 = 6561
log36561 = 8

b) 253/2 = 125
log25125 = 3/2

Evaluate:
a) log243/2
log2(22)3/2
log223

b) 5log510 = 10



Solve:

logt4 = 2
22 = t2
2 = t


Expand

log2 (m3 n5)1/2 = 3/2 log2m+ 5/2 log2n
bring down the 1/2, therefore

1/2 log2 (m3 n5)

Remember "power of a power"

log2 (m3/2 n5/2)
multiply the exponents
log2m3/2 + log2n5/2


Given: log5 = 0.70 log2 = 0.30
without using a calculator find log 0.08

Start of by saying 0.08 = 8/100
If you look at the number it can be simplified to 2/25

You must use a power to get a power which will get you 2/ 52

Answer: log 0.08 = log (2/52)

0.08 = log 2/52
= log 2 - 2 log5
= 0.30 - 2 (0.70)
= 0.30 - 1.40
= - 1.10

Solve: logax + loga(x-2) = loga3

logax(x-2) = loga3
x2-2x = 3
x2-2x-3 = 0
(x-3) (x+1) = 0
x=3 Accept
x=-1 Reject

We also had a talk about the population question in class. I am sorry guys because I lost the notes for it.


Journal Notes

EXPONENTS AND LOGARITHMS

Exponential Function: Any function where the variable is an exponent written in the form




f(x) = ab
THE ROLE OF PARAMETER A
a determines the y-intercept of the function a.k.a. the initial value (i.e. x=0) of the function.
a<0>the graph is reflected in the y-axis
THE ROLE OF PARAMETER B
b is the base of the exponential function
b is also known as the multiplication factor.
b>1 the function is increasing. This is known as "exponential growth".
0the function is decreasing. This is known as "exponential decay".
LOGARITHMS
Definition: 1) A logarithm is a function that turns a power into an exponent.
2) The logarithmic function, logba = c, is the inverse of the exponential function bc = a where b does not equal zero and a>0.
logb a = c
b is the base
a is the argument
c is the logarithm
which means
bc = a
b is the base
a is the power
c is the exponent
Examples: log381 = 4 means 34 = 81
log232 = 5 means 25 = 32
log4 (1/16) = -2 means 4-2 = 1/16
log51/125 = -3 means 5-3 = 1/125
log100 = 2 means 102 = 100

GRAPHICALLY
Since the logarithm function is the inverse of the exponential function...






Remember: A logarithm is an exponent!






LAWS OF LOGARITHMS
Product law: logamn = logam + logan
Example: log2(4 * 32) = log24 + log232

Quotient law: loga(m/n) = logam - logan
Example: log3(81/27) = log381 - log327

Power law: logamb = b logam
Example: log5253 = 3 log525

Change of base law: logam = logbm all over log ba
Example: log84 = log24 all over log28


SPECIAL CASES:

1) logaa = 1
2) loga1 = 0
3) logaax = x
4) alog a x = x

The Common Logarithm
A logarithm function to the base 10 ( log10x) is called a "common logarithm". This base is so frequently used that it is simply written as log x, without the base indicated.
When no base is indicated, base 10 is assumed.

THE NATURAL LOGARITHM

e = 2.718281828459...

e is a number that arises naturally in the study of exponential functions, particularly in the case of continuous exponential growth. The natural logarithm function is written as:

lnx i.e. logex = lnx

read as "el-en of x"

And thats the whole bottle of wax for today!

So I will end off with a joke, haha just joking guys. So the next scribe has to be jan. It was inevitable!



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blogging logs..

well this unit went by really fast.. and just like always theres always a test after. this unit is kind of complicating for me. i get mixed up with whats the power, base, exponent and so on. I'm not having to much trouble when it comes to the product rule, quotient rule and power law, but i do get confussed when there are alot of logs involved in the problem. when it comes to everything else i guess i'm okay with it.



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Tuesday, May 23, 2006

MR. BOB for LOGS AND EXPONENTS

this unit started goood, the problems we did in the first few days were simple to understand. I didn't find the problems difficult. Today's topic or section we looked over was a bit confusing. Maybe it was "that i forgot what a log is", lol but i do not think it was that. It was just the way that the log was used and exponent, until i saw the pattern. This unit was fun, because it gives me a different way to think and there are many ways to solve an equation. I find that is pretty similar to trig identities, the way that problems should be solved. A little more practise on the problems we did today and maybe using them in word problems or something and im good to go.



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Logarithms; What Fun

Can you sense my sarcasm? Oh, logarithms aren't so bad, really. I mean no sarcasm. We just need practice right? Well, what's a better way to start the week, than with a one period math class? We started class with eight questions given to us on the board; all to solve for x.

a) 2x = 3
b) 43x = 2
c) ux = v
d) a = bcx
e) log5x = -2
f) log3x = π
g) y = log43x
h) v = logbcx

ANSWERS:

a) x = log23 OR log3/log2

How we get log3/log2 :

2x = 3
log2x = log3
xlog2 = log3
x = log3/log2 [divide each side by log2]

b)The way I solved this problem was to change 4 as a power of 2.

22(3x) = 2
2(3x) = 1
6x = 1
x = 1/6

Mr. K, then showed us other ways to solve b.

log42 = 3x
(1/3)log42 = 3x(1/3)
(1/3)log42 = x
log42(1/3) = x
log43√2 = x

OR

43x = 2
log43x = log2
3xlog4 = log2
(1/3)3x = log2/log4(1/3)
x = log2/3log4
x = log2/log43

c) x = loguv OR logv/logu [it's exactly like question a), but with letters instead of numbers]

d) a/b = cx
logc(a/b) = x
logca - logcb = x

e) log5x = -2
x = 5-2
x = 1/25

f) x = 3π

g) y = log43x
log43x = y [I just switch sides because it seems less confusing for me with the x on the left side]
3x = 4y
x = 4y/3

h) logbcx = v
cx = bv
x = bv/c

So then we were given 3 questions to simplify:

a) 7log74
b) 2log25
c) 4log2[2(log25)]

So first, we always have to remember a logarithm is an exponent. Let's use a different example. If given log232, we'd know that the exponent is 5. 32 would be the result of 2(our base number) to the power of an exponent, and in this case it would be 5.

25 = 32

Then we can say that log232 is just another way to say 5. So if we were given 2log232 , it would be like saying 25, which is 32.


2log232 = 25

= 32

So if given different numbers, like 3log35, it would be stating that 3 to the power of a certain number, will result in 5.

3x = 5

So, 3x is a certain number itself, which is equivalent to log35 OR x, and will result in getting 5 in this given question.

If 3x = 5 then log35 = x THEREFORE 3log35 = 5 is the same as 3x = 5

I think that's how it goes. I hope I explained that right, but i think I may have just confused you guys even more. If that's still a bit vague, you can always just distiguish the pattern. If the base number is equilvalent to the base number attached to log, then the bigger number [sorry, I don't know the proper terminology], will be your answer.

alogab = b

EX. 6log68 = 8

So, the answers would be:


a) 7log74 = 4
b) 2log25 = 5

c) 4log2[2(log25)] = 4log25 [2log25 is just 5, using the pattern we

identified]

= (22)(log25) [we change 4 as a power of 2]

=(22log25)2

=2log225

=25

After that, we were given:

CHANGE OF BASE LAW

logam = logcm/logca

EX. log37 = log7/log3

=ln7/ln3

Mr. K then gave us two questions to work on, but we only had time to answer one.

1. log (x+6) + log(x-6) = 2

log (x+6)(x-6) = 2

log(x2-36) = 2

102 = x2-36 [I'm not 100% sure if that's how they relate or not]

100 = x2-36

0 = x2-136

0 = (x-√136)(x+√136)

x = √136 [accept] AND x = -√136 [reject]

2. [We didn't get to finish this question]

1/2loga(x+2) + 1/2loga(x-2) = 2/3loga27

Our homework is seven questions that's posted on the blog. Yay, I'm done my scribe! Marc, you're the next scribe. Have fun with covering two classes. =)






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Monday, May 22, 2006

blogging logarhythms

Ok sorry for the late blog. First things first using logs remember these

Use this with the quotient questions
Loga(m/n)=logam - logan

Use this with the multiplication questions
Loga(mn)= logam+logan

This one I'm not sure when to use it but it's tricky
cLogam=logama


express as a logarhythm of a single number or expression
a)loga2+ loga3
Loga(mn)
=loga(2*3)

b)loga5-loga4
Loga(m/n)
=loga(5/4)

c)3logLoga2
logama
loga32

d)-Loga(1/6)
logama
loga(1/6)-1

e)1/2logbm + 1/2 logbn
Loga(mn)
There are two ways to do this you can either put it as logbÖ(mn) or
Logbm1/2n1/2
=logbmn1/2

f)1/2(logbx-logb4)
Loga(m/n)
=logbÖ(x/4)


for these ones let C=Log3(10) and let d=Log3(5) and express in terms of C and D which means that when you simplify it it and it looks like either C or D then input it in for example Log3(10)+Log3(5) then replace it and you get C+D if you don't get it I'm sorry I don't really explain well but here are the questions you'll get and answers you'll see what I mean...

a)Log3250
=log3(52*10)
=log352+log310
=2log35+log310
+2D+C

b)log3log2
=log310/5
=log310-log35
=C-D

c)log3Ö(5)
=log3(5)1/2
=1/2log35

d)log31/100
=-2log310
=log3(1/1002)
=log31-log3102



it just says to expand so these questions
a)log2m5n4
=5log2m+4log2n

b)log2(mn)3
=3log2m+3log2n

c)log2 3Ö(m2n)
2/2log2m+1/3log2n

d)log2 (Ö(m/n3))
=1/2log2m-3/2log2n



solve these questions plus mr.K used something called the anti-log as shown on question A it gets rid of the log
a)logax=logaa+loga5
=anti-loga(logax)=anti-loga(loga45)
x=45

b)logax=loga 3+loga5
=logax=loga 3Ö(9)2+loga2
=anti-logax=anti-loga 3Ö(92)+ loga2
=log54
x=54

c)logb(x-+3)=logb8-logb2
=anti-logb(x-+3)=anti-logb8-logb2
x+3=4
x=1

This is where my notes get messy and I'm really unsure which is question and which is answer so I'll put it down as I see it.

d)logb(x2+7)=2/3logb64
logb(x2+7)=2/3logb64(2/3)
anti-logb(x2+7)=anti-logb16
x2+7=16
x2=9
x=-3, 3

Also another note you can check your answer by filling in for x

e)logbx-logb(x-5)=loga
logb(x/(x-5))=loga
anti-logb(x/(x-5))=anti-loga
x=6(x-5)
x=6x-30
x=5

f)loga(3x+5)-loga(x+5)=loga8
anti-loga(3x+5)-anti-loga(x+5)=anti-loga8
2x+5=8x+40
-35=5x
-7=x

I'm not sure if these are from the questions he gave us I just can't remember, but from what I see I think you would have to change the whole number into a logarthym and then you use the anti-log to find x

a)log2(x2)=4
log2(x2)=log216
anti-log2(x2)=anti-log216
x=-5, 5

b)log36(x+2)+log36=3
log3(6x+12)=3
log3(6x+12)=log327
anti-log3(6x+12)=anti-log327
x=5/2

c)log6(x+1)+log6=1
log6(x+1)+log6x=log01
anti-log6(x+1)+anti-log6x=anti-log01
\ no solution

Sorry for the long wait I just felt like doing the blog on a monday well it's a tuesday now but I started on monday If you see any mistakes feel free to tell me cuz I did it pretty late there's bound to be lot's of mistakes though, especially the last question. For the scribe umm... does anyone remember beat it by micheal jackson remember near the end where everyone crowded these two thugs and those two thugs had a weapon in one hand while the free hand was holding on to the other opponets free hand and the game was to hurt the other guy... well since there is marc and jessica left I believe they should do that to see who would be the scribe...winner doesn't have to do be scribe



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Wednesday, May 17, 2006

Lemon square!!! Scribe post!!!

Hi! guyzzz... We started the day with our terrifying test(just kidding). And finally Mr.K brought the lemon square, by the way its good Mr.K and can I have the recipe for that one.


In the afternoon, we started an example;

Evaluate the expression:
1.)


2.

3.

Then,Mark asked what if the example were given like this:

a.) 32 + 35

b.) 27 - 23

Mr.K , explained that there are no such rules that can fit in these such question because the first 2 example are implied within the property of multiplication and division.

Remember this following rules:


There is also an example where the 2 rules are join together like:


Well, I guessed that's it for me, by the way our homework is on page 106, numbers 1-42 only the even numbers. The next scribe will be Lerwyne.



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Tuesday, May 16, 2006

scribe post (post # 200)


although i didn't exactly finish this post as the 200th post, based on the post arrangement i'm still the 200th blogger here! well, i'm aldridge and i'm the scribe for today. i'm not looking forward joining the scribe's hall of fame for this scribe post cause this topic is really confusing to me and i don't get some of it. well today is a 1 period class. everybody seems to be earlier than the bell today and most of the class is passing around 3 bubble teas to be tasted. the bell rang twice and after a few moment Abdi enters the door. seems like no one notice him cause everyone is busy doing something except for Van who immediately put Abdi in the board after noticing him. Hopefully, Mr. K will bring the lemon squares tomorrow.
We did some review about probability today. the results form our quizstar quiz are already in. Mr. K discuss to us all of the question from that quiz. here they are:



these are the top 3 questions where most of us who take the quiz got it wrong. just click on the image to make it bigger.
here are the other 7 questions with their answers:


out of 16 students who do the quiz, the class got a total average mark of 72%, not bad for this unit.

sorry if this post took me long to post up. some of the statements I made here are already past tense but since the date posted is still may 16, its better to leave those statements in present and future tense. the next scribe will be emile.
I hope this is my last scribe post. haha... jk..



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Homework Assignment for May 16

Here is your homework assignment for tonight (20 Questions):

Questions Part 1 of 2
Questions Part 2 of 2

Here are the answers to the assignment:

Answers Part 1 of 3
Answers Part 2 of 3
Answers Part 3 of 3

This is post number 199 on our blog. Whoever posts next will write post number 200. I wonder what it will be? ;-)



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Monday, May 15, 2006

Scribe#55 : Review and Brief Introduction To Logirhythm

Okay, just because I wanted to look good, I spend 5 minutes counting what number scribe we're on. Apparently its 55. That's pretty cool. Anyways, I'm Van, and I'm gonna do one kick @55 scribe. (Sorry if that's offensive Mr. K).

Today was a 2 period day. Mr. K was lonely at the front. People slowly etched away, especially Sissyphus (That has to be a typo...). No one felt like moving, so Mr. K went on with the class. ALSO, Mr. K forgot about our lemon squares! The 80% probability failed. We gotta get em tomorrow!, hopefully, if Mr.K sees this scribe in time, he'll remember to make them right away.

So Mr. K posted the online quiz a LITTLE late, so it should end tonight at 11:59:59. So, get those done people! Mr. K mentioned about, all the BoB's(hehe I invented that) state that, most people have trouble with the poker probability. So we started doing 3 practice problems.

1. Find the probability of getting... *BOOM* *Mr.K turns around and looks at Calvin*

Calvin: "SO CLOSE!, I swear, that hole puncher was there as a trap."

Calvin failed to sneak into the room late while Mr.K wasn't looking. Now he owes him 3 bubble teas. Just thought I'd mention that.

1. Find the probability of getting 2 pair in a 5 card poker hand

A two pair is two cards of one rank and another two cards of another rank.

Ex.




















2. Find the probability of getting exactly 5 spades and 4 hearts in a 13 card bridge hand.

Okay, so what in the heck is bridge? I have no clue. I've never seen anyone play it, so let's post the rules for the heck of it, in case anyone wants to try.

http://www.pagat.com/boston/bridge.html








Find the probability of getting a "Full house" in a 5 card poker hand.

A full house, or full boat, contains a set (three) cards of one rank and a pair of another rank.

















So those were the practice questions, and the test will be on Wednesday MORNING. Get ready guys.

Sketch Each Pair Of Graphs On The Same Cartesian Plane

Mr. K: Start by gratch...

Mr. K said gratch. Definition: Mr. K's mistake of a new word. Appears to be a combination of "graph" and "sketch"



































So, I probably should've wrote what he was trying to say about exponents and powers. I'll update this again and ask him about the language. For now, here's the questions and answers

All I know is that, whatever the log base(under the log), the exponent(the # after =), is what I'm looking for, to make the power(beside the log)





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blogging

so yea I think I got it, I said I think I get it I'm not exactly sure if I trully got it, but still I understand it alittle bit but if it came down to the test I'll probably forget it hahaha. The problems I don't understand is the word problems involving the tree, the words like trick you sometimes just when I get it, it turns this corner and cofuses me hahaha. ohh and sorry bout the diffirent coulors I jsut felt bored



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blog post

well, our unit, "probability" is really hard for me! this is the hardest unit for me now! probably because i already forgot about permutations, combinations and binomial theorem! i think i need to review again those parts in our past unit. this probability thing should be easy for me cause i already took this at my statistics class and algebra(just for fun exercises in our class). i'm still trying to remember the easier way for doing word problems in this unit that i learned about two years ago. i hope i can remember it before our test cause i guess the probability that i'll pass this test is low. :-(



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Sunday, May 14, 2006

blogging for the test!

This unit was pretty short and way easier than the last unit. The hardest thing I would say about this unit is the questions dealing with a standard deck of cards. It just so hard to get my head to understand all those numbers in the deck. The easiest thing I found in this unit is using the tree diagrams to help solve the problem, it really helped. Well that's about it, oh and good luck everyone for the test!



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blogging on blog

I actually don't know what the test is about but, from what we do in class, it's seems pretty challenging. it's hard to grasp. When I first read the problem, it's just words and takes a while to actually get what the question is asking. Like what to use, the Choose thing or something else, what do first, what to do last. it's confusing. =S



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blogging on blogging

This unit is sort of easy. There is just some trouble when it comes to organizing the problem. For example when you have to find the sample space. I just did the online quiz and I started to understand how to do the questions. Another place that i have trouble on is when the notation GIVEN THAT is involved. I guess i just have to be more open minded.



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Blogging for probabilty on the blog

My blog is kinda late, but here it is. This unit realy does seem similar to the previos unit, counting. I would also say that this unit a little more easier compared to counting, because we do not have to deal with circles. This unit also helped me to better understand the use of the compliment. Some questions are are really tough to do, but it's fun since they are challenging. I have to be careful for the wording of some questions, because we have learned that "and" and "or" mean different things for probabilty questions. Other than that i don't think i had much problem with this unit. Again, practise is the key, i think to understand this unit and really diggin deep into what a problem asks for.



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Saturday, May 13, 2006

Probability Online Review Quiz

I had some trouble with Quizstar yesterday. The quiz went online this morning. I've extended the deadline to midnight Monday night. The unit test will be on Tuesday or Wednesday. We'll talk about it in class on Monday.

You can write the quiz here. You have 30 minutes to complete it. If you haven't yet registered at QuizStar then when you get to the site follow these instructions:

  1. Click on the big yellow "Sign Up" arrow.

  2. Use only your first name and last initial as indicated.

  3. Pick a username that will allow me to easily identify you, i.e. first name and last initial.

  4. Make up any password you like.

  5. Click on [Register] then [Search] by teacher's name (kuropatwa) and you'll find me.

  6. Click on the box next to Pre-Cal 40S and then [Register].

  7. Follow the instructions on the screen.


Actually, if you read each page carefully, you'll see that the sign up process is very straight forward and self explanatory. If you hit any snags email me and we'll sort it out together.

Remember, the quiz is timed. You'll only have 30 minutes to complete it once you've begun. It consists of 10 multiple choice questions (1 mark each). DON'T PANIC. Take your time. I know you'll all do well. ;-)

Do Your Best!



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Friday, May 12, 2006

OMG 10 AMAZING SCRIBES IN A ROWW



Oh my god Jacky that was an AWSOME scribe and thanx for choosing me the lucky scribe for today:) I would also like to say to the last 9 scribes in a row they did an amazing and an out standing job on there scribes.
Hay class mates i hope that you are starting to study for this units TEST. This unit was short unlike the other ones. Okay today was a one period class and as always it starts at 1:20 pm and if you were not in Mr.K class by that time you must bey him Ice Tea. We got alot of flavours covered alrady but we want to cover them ALL so keep on working on that class mates :) And today we just did a pre-test.

The pre-test is as fallows
Calculators are allowed.

Question:
1) Twelve people, including you, are members of a choir. The choir director is going to choose three members to attend a workshop. The probability you and two other members will be chosen is:
a) 1/4 b)3/10 c)1/12 d)1/10
Solution: Ther are 3 to be chosen out of 12 people 3/12 reduce it = 1/4 Therefore, the answere you must choose is (a).

Question:
2) Rex is playing a guessing game. The probability he will guss each question correctis 0.3. What is the probability he will guess exactly 5 out of 10 questions correct?
a)0.15 b)0.10 c)0.29 d)0.80

Solution: (C)= Correct answers (W)= Wrong answers p(c)=0.3 CCCCC WWWWW (10C5)(0.3)^8(0.7)^5 =0.10 Therfore, the one you choose should of been (b)
Question:
3) The serial number of a $10 bill contains 8 digits. If your $10 bill contains the digit 7 at least once, you win a prize. What is the probability that you'r $10 bill will win?






Long Answer Questions
4) Jhon takes the bus to school on two days, and on the other three days he walks. If he takes the bus he is late 10% of the time, but if he walks he is late 30% of the time.
a) What is the probability that John arrives on time?



Solution: On time:p(Bo)+p(Wo) =0.36+0.42 =0.78 (100) =78%


b) If jhon was late, what is the probability that he took the bus? solution:
P(Bl/l)= p(Bl)/p(Bl)+P(Wl) =0.04/(0.04)+(0.18) =(0.4)/(0.22) =0.1818 (100) =18%

And the Next Lucky scribe is .......i need a drum roll :) I can't dacide so i put the names in a hat and i picked one out and the next scribe is Van. It was the only way i could be fair and after all this is the end of the probability unit. :P :P



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